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Lets say we have $\triangle$$ABC$ having $O,I,H$ as its circumcenter, incenter and orthocenter. How can I go on finding the area of the $\triangle$$HOI$.

I thought of doing the question using the distance (length) between $HO$,$HI$ and $OI$ and then using the Heron's formula, but that has made the calculation very much complicated. Is there any simple way to crack the problem?

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5 Answers 5

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Hint:

$AH=2R\cos A$, $AI=4R\sin\frac{B}{2}\sin\frac{C}{2}$, $OA=R$, $\angle OAH=2\angle OAI=B-C$

$\triangle HOI=\triangle AHO-\triangle AHI-\triangle AIO$

Final answer $$2R^2\sin\dfrac{B-C}{2}\sin\dfrac{C-A}{2}\sin\dfrac{A-B}{2}$$

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Given any triangle $\triangle ABC$, we will abuse notation and use the same letter to represent both a vertex and the angle at that vertex. Let

  • $a, b, c$ be the side lengths $|BC|$, $|CA|$ and $|AB|$.
  • $L = a + b + c$ be the perimeter.
  • $c_X, s_X, t_X$ be $\cos X, \sin X, \tan X$ for any angle $X \in \{ A, B, C \}$.
  • $R$ be the circumradius.

Let $I, O, H$ be the incenter, circumcenter and orthocenter.
Their barycentric coordinates are given by

$$\begin{array}{ccccc} \alpha_I : \beta_I : \gamma_I &=& \sin A : \sin B : \sin C &=& t_A c_A : t_B c_B : t_C c_C\\ \alpha_O : \beta_O : \gamma_O &=& \sin 2A : \sin 2B : \sin 2C &=& 2t_A c_A^2 : 2t_B c_B^2 : 2t_C c_C^2\\ \alpha_H : \beta_H : \gamma_H &=& \tan A : \tan B : \tan C &=& t_A : t_B : t_C \end{array}$$

Let $\mathcal{A}_0$ and $\mathcal{A}$ be the area of $\triangle ABC$ and $\triangle IOH$, their ratio is given by

$$\frac{\mathcal{A}}{\mathcal{A}_0} = \left|\det\begin{bmatrix} \alpha_I & \beta_I & \gamma_I\\ \alpha_O & \beta_O & \gamma_O\\ \alpha_H & \beta_H & \gamma_H \end{bmatrix}\right| = \frac{\mathcal{N}}{\delta_I\delta_O\delta_H} $$ where $\;\begin{cases} \delta_I &= \sin A + \sin B + \sin C\\ \delta_O &= \sin 2A + \sin 2B + \sin 2C\\ \delta_H &= \tan A + \tan B + \tan C \end{cases} \;$ and $$\begin{align}\mathcal{N} &= \left|\det\begin{bmatrix} t_A c_A & t_B c_B & t_C c_C\\ 2t_A c_A^2 & 2t_B c_B^2 & 2 t_C c_C^2\\ t_A & t_B & t_C \end{bmatrix}\right| = 2t_A t_B t_C \left|\det\begin{bmatrix} c_A & c_B & c_C\\ c_A^2 & c_B^2 & c_C^2\\ 1 & 1 & 1 \end{bmatrix}\right|\\ &= 2t_A t_B t_C |(c_A - c_B)(c_B - c_C)(c_C - c_A)| \end{align} $$ Since $A + B + C = \pi$, $\delta_H = t_A + t_B + t_C = t_A t_B t_C$. Together with the relations, $$\begin{cases} \mathcal{A}_0 &= \frac12 R^2(\sin 2A + \sin 2B + \sin2C)\\ L &= 2R(\sin A + \sin B + \sin C) \end{cases} \quad\implies\quad \begin{cases} \delta_O = \frac{2\mathcal{A}_0}{R^2}\\ \delta_I = \frac{L}{2R} \end{cases} $$ We find $$\frac{\mathcal{A}}{\mathcal{A}_0} = \frac{2}{\delta_I\delta_O}|(c_A - c_B)(c_B - c_C)(c_C - c_A)| = \frac{2R^3}{\mathcal{A}_0L}|(c_A - c_B)(c_B - c_C)(c_C - c_A)| $$ Notice $$c_A - c_B = \frac{-a^2 + b^2 + c^2}{2bc} - \frac{a^2-b^2+c^2}{2ac} = \frac{(b-a)L(L-2a)}{2abc}$$ and similar expressions for $c_B - c_C, c_C - c_A$, we find $$\frac{\mathcal{A}}{\mathcal{A}_0} = \left(\frac{2R^3}{\mathcal{A}_0L}\cdot\frac{L^3(L-2a)(L-2b)(L-2c)}{8a^3b^3c^3}\right)|(a-b)(b-c)(c-a)| $$ Recall the Heron's formula and a beautiful relation between $\mathcal{A}_0$ and $R$: $$16\mathcal{A}_0^2 = L(L-2a)(L-2b)(L-2c)\quad\text{ and }\quad 4\mathcal{A}_0 R = abc $$ What's inside the parenthesis above can be simplified as

$$\frac{2R^3}{\mathcal{A_0}L}\cdot\frac{L^2\cdot16\mathcal{A_0}^2}{8(4\mathcal{A}_0R)^3} = \frac{L}{16A_0^2} = \frac{1}{(L-2a)(L-2b)(L-2c)}$$ This leads to a reasonably simple ratio one can use to compute the area $\mathcal{A}$.

$$\frac{\mathcal{A}}{\mathcal{A}_0} = \frac{|(a-b)(b-c)(c-a)|}{(-a+b+c)(a-b+c)(a+b-c)}$$

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  • $\begingroup$ @JackD'Aurizio thanks for the compliment. $\endgroup$ Commented Jun 17, 2016 at 0:52
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From the information of incentre and Euler line, we have:

\begin{align*} r &= \frac{\Delta}{s} \\ R &= \frac{abc}{4\Delta} \\ IH &= \sqrt{4R^{2}+2r^{2}-\frac{1}{2}(a^2+b^2+c^2)} \\ IO &= \sqrt{R(R-2r)} \\ OH &= 9R^{2}-(a^2+b^2+c^2) \\ \end{align*}

It's do-able by using simple program in a computer. You can also find the area bound $(\Delta HOI=\frac{d}{2} \times OH)$ by knowing the distance bound between the incentre and the Euler line.

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If you compute the exact barycentric coordinates of $O,H,I$, the ratio $\frac{[OHI]}{[ABC]}$ is given by a simple determinant. So, just exploit the first table on this page and perform some computation.

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A área do triângulo HOI do triângulo ABC de lados a=|BC|, b=|AC| e c=|AB| é dada por: $AreaHOI=|\frac{1}{4}.\frac{\left|\begin{array}{} a & a^3 & 1 \\ b & b^3 & 1 \\ c & c^3 & 1 \\ \end{array} \right| }{\left|\begin{array}{} 0 & a^2 & b^2 &1\\ a^2 & 0 & c^2&1 \\ b^2 & c^2 & 0&1 \\1&1&1&0 \end{array} \right| }|$

Dedução: Seja o triângulo ABC, com o vértice A na origem do Sistema Cartesiano e C no semi eixo positivo x. Assim:

$A=(0,0)$

$B=(\frac{-a^2+b^2+c^2}{2b},\frac{2S}{b})$

$C=(b,0)$

$S=\sqrt{p(p-a)(p-b)(p-c)}$

$p=\frac{a+b+c}{2}$

$AreaHOI=3.AreaGIO$

$AreaGIO=\frac{1}{2}\left|\begin{array}{} xG & yG & 1 \\ xI & yI & 1 \\ xO & yO & 1 \\ \end{array} \right| $

"Obtendo as coordenadas de G, I e O através de suas respectivas fórmulas e fazendo substituições obtemos:"

$AreaHOI=3.\frac{1}{2}\left|\begin{array}{} \frac{-a^2+3b^2+c^2}{6b} & \frac{2S}{3b} & 1 \\ \frac{-a+b+c}{2} & \frac{2S}{a+b+c} & 1 \\ \frac{b}{2} & \frac{b(a^2-b^2+c^2)}{8S} & 1 \\ \end{array} \right| $

"Desenvolvendo, substituindo S, p e fatorando obtemos:"

$AreaHOI=\left|\frac{(a-b)(a-c)(b-c)(a+b+c)}{16S} \right| $

"Substituindo as expressões por determinantes (S no formato de área de Cayley-Menger), obtemos:"

$AreaHOI=|\frac{1}{4}.\frac{\left|\begin{array}{} a & a^3 & 1 \\ b & b^3 & 1 \\ c & c^3 & 1 \\ \end{array} \right| }{\left|\begin{array}{} 0 & a^2 & b^2 &1\\ a^2 & 0 & c^2&1 \\ b^2 & c^2 & 0&1 \\1&1&1&0 \end{array} \right| }|$

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